What is the vertices y^2+12x-2y-11=0 ?

The vertices y^2+12x-2y-11=0 is (h,k)=(1,1),p=-3

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vertices y^2+12x-2y-11=0

vertices

y^{2} + 12 x - 2 y - 11 = 0

(h, k) = (1, 1), p = - 3

Solution steps

y^{2} + 12 x - 2 y - 11 = 0

Rewrite

y^{2} + 12 x - 2 y - 11 = 0

in the standard form

4 (- 3) (x - 1) = (y - 1)^{2}

Therefore parabola properties are:

(h, k) = (1, 1), p = - 3

f (x) = x^{2} + 5 x - 2

x^{2} + 6 x + 3 y + 2 = 0

y = - x^{2} + 8 x - 7

- 3 x^{2} + 6 x + 5

f (x) = - x^{2} + 12 x - 35

What is the vertices y^2+12x-2y-11=0 ?
The vertices y^2+12x-2y-11=0 is (h,k)=(1,1),p=-3